need help solving this

Question

Find all the real zeros of the polynomial. Use the quadratic formula if necessary. (Enter your answers as a comma-separated list. Enter all answers including repetitions.)

P(x) = 2x3 + 8x2 + 6x − 4

Answer 1

2x^3 +8x^2 +6x -4

use the cubic formula, although it's complicated and not easy to use. It's relatively unknown

or graph and find the x intercepts=roots=zeros

= about -2.41421, -2, and 0.41421

= exactly -1-sqr2, -2, and -1+sqr2

3 real zeros

every odd degree polynomial has at least 1 real zero

1, 2 or 3 real zeros

same zeros as

x^3 +4x^2 +3x -2

= (x+2)(x^2 +2x-1)

set each factor =0 and solve for x

x =-2

and

x^2 +2x -1 = 0

use the quadratic formula for 2 more zeros

x = -2/2 +/-.5sqr(4 +4)/2 = -1 +/-sqr2 = 1+sqr2 or 1-sqr2

= about -2.1416 and 0.1416

Answer 2

A minor solution strategy:

2x³ +8x² + 6x -4 = 0

Since we can factor out 2, and divide both sides by 2, the solution set of the above equation is identical to that of:

x³ + 4x² + 3x -2 = 0

This makes applying the rational root theorem somewhat simpler:

±{1, 2} is the entire set of potential rational solutions.

With the original equation, the entire set of potential rational solutions would be:

±{1, 2, 1/2, 4}.

In essence, we cut our solution search in half.

And -2 turns out to be a solution, so we can factor that out and reduce a cubic equation to a quadratic one. Applying the quadratic formula provides us with a pair of irrational "conjugate" roots.

And that is all three roots of this cubic equation.

Answer 3

-2, -1 - √2, -1 + √2